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February 1, 2017, 21:25 
conjugate heat transfer problem

#1 
New Member

Hi there,
I am using TEACH code for my project. To simulate a conducting wall inside a close cavity I faced a strange problem. What I did: 1 I wrote the code for natural convection in a cavity 2 Defined a vertical wall in the cavity and assigned zero velocity for those cells 3 For energy equation I changed diffusion coefficient from (1/(Ra.Pr)^0.5) to (Kr/(Ra.Pr)^0.5) which Kr is conductivity coefficient of wall over fluid. (dimensionless form of governing equation are shown in the picture.) I expected code to solve fluid part and when reaches to solid section conductivity changes and only conduction should be in the wall. However, interestingly the coefficient I add does not change thermal conductivity of wall and results show that conductivitis of wall and fluid are same. Any body knows how to fix it? Patankar explains it in his book (page 149) but I do not understand it. Any help appreciated. 

February 2, 2017, 03:29 

#2 
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Filippo Maria Denaro
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I do not understand your question, you solve for the fluid and the solid zone two different temperature equations, each one with different diffusion coefficient (if Kr is not 1), isn't that? Now what you are checking in this two zones? could you plot the fields?


February 2, 2017, 03:38 

#3 
New Member

I am studying partitioned cavity as showed in the following figure:
As you can see in the above figure, although I changed thermal conductivity of the partition in the middle of the cavity, it shows same thermal conductivity as fluid. If thermal conductivity was different, we should see breaking point for isotherms at the interface of partitionfluid. Now I am not should what I did wrong. Should I change more or just changing diffusion coefficient and making velocities zero in the partition is enough? Since usually most of people are familiar with TEACH code, I was hopping people help me with the items I should change to be able to have functioning conducting partition in the cavity. 

February 2, 2017, 03:53 

#4 
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Filippo Maria Denaro
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At the wall you prescribe the continuity of the heat flux, that is q_wall = K grad T, right? Now, what about the value of Kr? if it is of O(1) the difference is disregardable.


February 2, 2017, 04:00 

#5 
New Member

Oh, thanks god you are here...
We should apply the continuity of flux at the interface but Patankar says instead we can increase viscosity of solid section to make velocity zero. For the energy equation, he says we can set solid property in the energy equation and solve whole domain without worrying about continuity. Now I did that but I am not getting big difference for partition section as I assumed Kr=1000 !! Should I also apply that constant heat flux for interface ? or other thing ? How ? 

February 2, 2017, 04:06 

#6  
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Filippo Maria Denaro
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Quote:
yes, you must apply the continuity of the flux at the wall and, simultaneously, setting zero velocity for the solid zone 

February 2, 2017, 04:08 

#7 
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February 2, 2017, 04:19 

#8 
New Member

I did as follow but my code stopped working and got error:
Tw=T(w1)  (Kf/Kw)(dxf/dxw)(T(w2)T(w1)) Should it be something like above or I am totally in wrong direction ? 

February 2, 2017, 04:22 

#9 
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Filippo Maria Denaro
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what kind of error? just after one time step?


February 2, 2017, 04:25 

#10 
New Member

It stopped working. Even not one step. I am guessing, I do not know correct way of making sure of flux continuity.
Could you please give me some details about how I can write between partition and fluid cells ? Much appreciated 

February 2, 2017, 04:28 

#11 
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Filippo Maria Denaro
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q=k dTdn, you discretize the RHS with a backwar/forward second order derivative and consider that q = h* Delta T0


February 2, 2017, 04:33 

#12  
New Member

Quote:
(k dTdn )fluid = (k dTdn )solid so I have two cells from fluid side for left equation and two cells from solid side for right equation. The question is how can I write it for code to be able to solve it ? 

February 2, 2017, 04:41 

#13  
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Filippo Maria Denaro
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Quote:
no, the flux in the solid is q = h *(TTr), not the Fourier flux 

February 2, 2017, 04:47 

#14  
New Member

Quote:
and I have checked all the papers in this regard and they all say this : as a boundary condition for conducting partition. Should I write this continuity flux for both of partition ? 

February 2, 2017, 04:53 

#15 
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Filippo Maria Denaro
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the value of h depends on the material of the solid zone


February 2, 2017, 05:14 

#16 
New Member

Do you mean something like the picture ?
What is Tr in your note ? (Ts  ? ) Should I do this calculation for both sides ? For the left side I am going to calculate Ts (as in picture) what about right side ? T(E+1) ? 

February 2, 2017, 05:41 

#17 
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Filippo Maria Denaro
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yes, correct, the expression of q is well known, see for example here
https://en.wikipedia.org/wiki/Convective_heat_transfer you have the difference between fluid and wall temperature 

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coding, conjugate heat transfer, partition 
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